Saturday, November 28, 2015

leetcode面试题2 抄书问题


leetcode面试题2 抄书问题
http://codevs.cn/problem/3162/

有n本书和k个抄写员。要求n本书必须连续地分配给这k个抄写员抄写。也就是说前a1本书分给第一个抄写员,接下来a2本书分给第二个抄写员,如此类推(a1,a2需要你的算法来决定)。给定n,k和每本书的页数p1,p2..pn,假定每个抄写员速度一样(每分钟1页),k个抄写员同时开始抄写,问最少需要多少时间能够将所有书全部抄写完工?(提示:本题有很多种算法可以在不同的时间复杂度下解决,需要尽可能的想到所有的方法)

解答

解法1:动态规划
设f[i][j]代表前i本书分给j个抄写员抄完的最少耗时。答案就是f[n][k]。状态转移方程f[i][j] = min{max(f[x][j-1], sum(x+1, i)), j<x<i}。其中x是在枚举第j个抄写员是从哪本书开始抄写。
时间复杂度O(n^2*k)
public int shortestTime(int[] pages, int k) {
int n = pages.length;
int[][] states = new int[k + 1][n + 1];
int[] sums = new int[n + 1];
sums[0] = 0;
sums[1] = pages[0];
states[1][1] = pages[0];
for (int i = 2; i <= n; i++) {
sums[i] = sums[i-1] + pages[i-1];
states[1][i] = sums[i];
//System.out.println(sums[i]);
}

for (int i = 2; i <= k; i++) {
for (int j = 1; j <= n; j++) {
int minVal = Integer.MAX_VALUE;
for (int jj = 0; jj <= j; jj++){
int val = Math.max(states[i-1][j - jj] , sums[j] - sums[j-jj]);
if (val < minVal){
minVal = val;
}
}
states[i][j] = minVal;
}
}
return states[k][n];
}

下面代码是DP:但是没有考虑边界情况只是写来熟悉算法流程。
http://www.cnblogs.com/cane/p/3894669.html
11 //dp:f[i][j]表示前i本书由j个抄书员来抄的最少时间,
12 //f[i][j] = min(f[x][j-1]+sum(x+1,i))(j<=x<=i)
13 ///sum[i][j]表示抄写第i到第j本书需要的时间
14 vector<int> book;
15 int dp[100][100];
16 int sum[100][100];
17 int fun(int n,int m)
18 {
19     int i,j,k;
20     int mintime=INT_MIN;
21     
22     for(i = 0 ; i < n ; ++i)
23     {
24         for(j = i ; j < m ;++j)
25         {
26             for(k = i ; k <= j ; ++k)
27             {
28                 sum[i][j]+=book[k];
29             }
30         }
31     }
32     for(i = 1 ; i <= n ; ++i)
33         dp[i][1] = sum[0][i-1];
34     for(j = 2 ; j <= m ; ++j)
35     {
36         for(i = j ; i <= n ; ++i)
37         {
38             dp[i][j]= INT_MAX;
39             for(k = j ; k <= i ; ++k)
40             {
41                 mintime = max(dp[k][j-1],sum[k][i-1]);
42                 dp[i][j] = min(dp[i][j],mintime);
43             }
44         }
45     }
46     return dp[n][m];
47 }
解法2;动态规划+决策单调。
同上一解法,但在x的枚举上进行优化,设s[i][j]为使得f[i][j]获得最优值的x是多少。有s[i][j+1]>=s[i][j]>=s[i-1][j]。因此x这一层的枚举不再是每次都是n而是总共加起来n。
时间复杂度O(n*k)

解法3:二分答案
二分答案,然后尝试一本本的加进来,加满了就给一个抄写员。看最后需要的抄写员数目是多余k个还是少于k个,然后来决定是将答案往上调整还是往下调整。
时间复杂度O( n log Sum(pi) )
http://blog.csdn.net/csyzcyj/article/details/17332821
这是经典的求最大值最小的问题,用二分答案。二分一个单人抄书的最大值,然后从后向前让每个人尽可能多抄<不多于二分的值>,若抄完了整本书,则下调上界<即可能人没轮完就取完了>,若还未抄完整本书就轮完了所有人,则上调下界<即还未取满就完了>,当上界=下界时退出,按同样的方法从后往前取书
int M,K,a[MAX],min1=IMAX,max1=0,x[MAX],y[MAX],ans;
bool check(int x)
{
      int sum=0,now=M;
      for(int i=K;i>=1;i--)
      {
            sum=0;
            while(sum+a[now]<=x && now>0)
            {
                  sum+=a[now];
                  now--;
            }
            if(now==0)   return true;
      }
      return false;
}
void work()
{
      int left=min1,right=max1;
      while(left<right)
      {
            int middle=(left+right)/2;
            if(check(middle))//说明最大值多了<即可能人没轮完就取完了>
                  right=middle;
            else left=middle+1;//说明最大值少了<即还未取满就完了> 
      }
      ans=left=right;
}
int main()
{
      //freopen("input.in","r",stdin);
   //freopen("output.out","w",stdout); 
   scanf("%d%d",&M,&K);
   for(int i=1;i<=M;i++)
   {
         scanf("%d",&a[i]);
         min1=min(a[i],min1);
         max1+=abs(a[i]);
      }
      work();
      
      int sum=0,now=M;
      for(int i=K;i>=1;i--)
      {
            sum=0;
            y[i]=now;
            while(sum+a[now]<=ans && now>=i)
            {
                  sum+=a[now];
                  now--;
            }
            x[i]=now+1;
      }
      for(int i=1;i<=K;i++)
            printf("%d %d\n",x[i],y[i]);
      return 0;
}

面试官角度:
该问题的考点在于算法能力。需要一定的算法积累,如对动态规划和二分法的算法积累。面试官不一定会需要你答出所有的解法(根据职位要求和招聘名额来看了),但是你答出多少就能够大概知道你在算法能力上的水平是多少。一般来讲至少需要答出动态规划的解法,因为只要稍微做过一点动态规划的训练,都是可以想出来的。

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